Surface Areas & Volumes | Exercise 13.3 | Q8

CBSE Class 9 Maths | NCERT Solutions to Exercise | Curved Surface Area of Cone

Question 8: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of the painting is Rs.12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)


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Explanatory Answer | Exercise 13.3 Question 8

Given Data: Height of the conical barricade, h = 1 m = 100 cm
Diameter of the conical barricade, d = 40 cm
∴ Radius of the conical barricade, r = \\frac{40}{2}) = 20 cm (or) 0.2 m

Step 1: Compute Slant Height of Cone

Square of slant height of the cone , l2 = h2 + r2
l2 = 202 + 1002
l2 = 400 + 10,000 = 10400
l2 = = 10,000 × 1.04
So, slant height, l = 100 (√ 1.04)
l = 100 × 1.02 = 102 cm (or) 1.02 m

Step 2: Compute Curved Surface Area of Cone and that of 50 Conical Barricades

Curved surface area of the conical barricade = π × r × l
Take π = \\frac{22}{7})
= 3.14 × 0.2 × 1.02
= 0.64056 m2

Curved surface area of 50 conical barricade = 0.64056 × 50
= 32.028 m2

Step 3: Compute Cost of Painting All Conical Barricades

Cost of painting 50 conical barricades at Rs.12 per m2 = 12 × 32.028
= Rs.384.336

 


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